Y -tan -1log e e - x 2-log e ex 2-tan -13-2 log e x -1-6 log X x - then d 2 y - dx 2 isA- 2B- -1C- 1Dh

The correct option is D 0y = tan^ 1 ( log_e(e/x^2)/log_e(ex^2) ) + tan^ 1 ( 3 + 2log_ex/1 6log_ex ) = tan^ 1 ( 1 2log_ex/1 + 2log_ex )+tan^ 1 ( 3 + 2log_ex/ ...

You visited us 7 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

y =tan -1log ...

Question

$I f y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}), t h e n \frac{d^{2} y}{d x^{2}} i s$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.0
$y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) = t a n^{- 1} (\frac{1 - 2 l o g_{e} x}{1 + 2 l o g_{e} x}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 3.2 l o g_{e} x}) = t a n^{- 1} (1) - t a n^{- 2} (2 l o g_{e} x) + t a n^{- 1} (3) + t a n^{- 1} (2 l o g_{e} x) = t a n^{- 1} (1) + t a n^{- 1} (3) \frac{d y}{d x} = 0$

Suggest Corrections

Similar questions

Q. If

y = T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) + T a n^{- 1} ⎛ ⎜ ⎝ \frac{l o g_{e} (\frac{e}{x^{2}})}{l o g_{e} (e x^{2})} ⎞ ⎟ ⎠, t h e n \frac{d^{2} y}{d x^{2}} i s

I f y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}), t h e n \frac{d^{2} y}{d x^{2}} i s

i f y = T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 + 6 l o g_{e} x}) + T a n^{- 1} ⎛ ⎜ ⎝ \frac{l o g_{e} (\frac{e}{x^{2}})}{l o g_{3} (e x^{2})} ⎞ ⎟ ⎠, t h e n \frac{d^{2} y}{d x^{2}} i s

Q. If

y = {tan}^{- 1} ⎛ ⎝ \frac{log \frac{e}{x^{3}}}{log (e x^{3})} ⎞ ⎠ + {tan}^{- 1} (\frac{3 + 3 log x}{1 - 9 log (x)})

,then

\frac{d^{2} y}{d x^{2}} =

y = {tan}^{- 1} ⎡ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎤ ⎥ ⎥ ⎦ + {tan}^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

, then

\frac{d^{2} y}{d x^{2}} =

Join BYJU'S Learning Program

If y=tan−1(loge(e/x2)loge(ex2))+tan−1(3+2logex1−6logex),then d2ydx2 is

The correct option is C 0.0y=tan−1(loge(e/x2)loge(ex2))+tan−1(3+2logex1−6logex)=tan−1(1−2logex1+2logex)+tan−1(3+2logex1−3.2logex)=tan−1(1)−tan−2(2 logex)+tan−1(3)+tan−1(2 logex)=tan−1(1)+tan−1(3)dydx=0

$I f y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}), t h e n \frac{d^{2} y}{d x^{2}} i s$